Find the sum of all positive integral values of n for which \(\frac{n+6}{n}\) is an integer.
n can be any multiple of 6 so the sum is infinite.
This answer is compeletely wrong so skip it.
Find the sum of all positive integral values of n for which
\(\mathbf{\large\dfrac{n+6}{n}}\)
is an integer.
\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{n+6}{n}} \\\\ &=& \dfrac{n}{n} + \dfrac{6}{n}\\\\ &=& 1 + \dfrac{6}{n}\\ \hline \end{array}\)
So \(n\) are the divisors of \(6\).
The 4 divisors of 6 are: \(1,2,3,6\)
The sum is \(1+2+3+6 =\mathbf{ 12}\)
Listfor(n, 1, 10, (n+6) mod n) =(0, 0, 0, 2, 1, 0, 6, 6, 6, 6). Only 4 numbers that give "0" as the remainder satisfy the condition, as heureka found. They are: 1 + 2 + 3 + 6 =12. After 6, all other numbers will give a remainder of 6.
Hi MobiusLoops,
I do not believe you are using the word correctly.
This is an Integer or integer value question.
Integrals are completely different. They are a part of calculus and nothing to do with integer values.
I understand the confusion, Integral is an normal English word and integral and integer would have the same origin in speech. But in maths the two meanings are very different.
Melody, I don't think so. My math question stated that question sooo... I don't know if they made a mistake or I'm not sure.